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Question

The vector from the point $$-\hat{i} + 2\hat {j} + 6\hat{k}$$ to the straight line through the point $$(2, 3, -4)$$ and parallel to the vector $$6\hat{i} + 3\hat {j} + 4\hat {  k}$$, is  


A
29761i+11861j53461k
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B
29761i11861j53461k
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C
29761i+11861j+53461k
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D
none of these
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Solution

The correct option is A $$\dfrac{297}{61}i+\dfrac{118}{61}j-\dfrac{534}{61}k$$

Equation of the straight line will be 
$$\dfrac{x-2}{6}=\dfrac{y-3}{3}=\dfrac{z+4}{4}=t$$
Therefore
$$x=6t+2$$
$$y=3t+3$$
$$z=4t-4$$
Therefore, let any point P on the line be 
$$P=(6t+2,3t+3,4t-4)$$
And $$A=(-1,2,6)$$
Therefore 
$$\vec{AP}=(6t+3)i+(3t+1)j+(4t-10)k$$
Now for perpendicular distance
$$\vec{AP}.(6i+3j+4k)=0$$
Or 
$$((6t+3)i+(3t+1)j+(4t-10)k).(6i+3j+4k)=0$$
$$6(6t+3)+3(3t+1)+4(4t-10)=0$$
$$36t+18+9t+3+16t-40=0$$
$$61t-19=0$$
$$t=\dfrac{19}{61}$$
Hence 
$$\vec{AP}=\dfrac{297}{61}i+\dfrac{118}{61}j-\dfrac{534}{61}k$$
Hence 
$$d=|\vec{AP}|$$

Mathematics

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